How to transform correctly an array to json_encode

Question

I found this and its very helpful...

Demo: jQuery Ajax Call to PHP Script with JSON Return

But i have some problem with to use it....

I need to understand how to use it correctly...

I dont understand why my array cannot be transform in a json....

These code are in response.php :

$champs = ["user" => "", "combinaison" => 0, "valeurSmall" => "", "valeurBig" => "", "aucune_valeur" => false, "trop_valeur" => false, "color_red" => 0, "color_green" => 0, "color_blue" => 0];

$return["json"] = json_encode($champs, JSON_FORCE_OBJECT); 
echo json_encode($return);

And in my index.html the part javascript :

$.ajax({
    type: "POST",
    dataType: "json",
    url: "changerMise/changer.php", 
    data: data,
    success: function(dataReturn) {                    
    alert("Form submitted successfully.\nReturned json: " + dataReturn["json"] );
       }
   });

The alert dont come back with the thing i need to use after....

The result gave :

Form submitted successfully.
Returned json: {"aucune_valeur": true}

any ideas ?!??!

Thx guys

echo json_encode($champs,JSON_FORCE_OBJECT); ...

YvesLeBorg
– YvesLeBorg

2018-08-13 02:36:52 +00:00
Commented Aug 13, 2018 at 2:36 — YvesLeBorg
– YvesLeBorg, Commented Aug 13, 2018 at 2:36

Dimas Lanjaka · Accepted Answer · 2018-08-13 02:06:23Z

0

Are you has give header json for response output on top of your php file, like below ?

<?php header("Content-type: application/json; charset=utf-8"); ?>

answered Aug 13, 2018 at 2:06

Dimas Lanjaka

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4 Comments

Dimas Lanjaka Over a year ago

If you using POST method you must know where data name/value will be executed/inspected. $_POST["json"] example, there data from inputed form will be executed/testing.

Dimas Lanjaka Over a year ago

You must add header json, and add allow cors origins

Dimas Lanjaka Over a year ago

Yes. Ajax json must be reading json object (not html). I think your PHP returned html header, not json.

Benoît Mignault Over a year ago

I know why i didnt receive any information at my output responce...At the end of one of my function, i forgot this ««««return $champs;»»» Please don't throw any rotten tomatoes lol

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